Austrian Mathematical Olympiad

Figure 2: Problem 6

Let $H_a$ be the foot of the altitude on $BC$. With the angle sum in triangle $A{H}_{a}C$, we get $$\angle HAC=90^{\circ }-\angle BCA.$$ Let $H_b$ be the foot of the altitude on $AC$. With the angle sum in triangle $C{H}_{b}B$, we get $$\angle CBH=90^{\circ }-\angle BCA.$$ The inscribed angle theorem gives us $$\angle CPH=\angle CBH,$$ therefore $$\angle CPH=\angle HAC.$$ We conclude that the triangle $AHP$ is isosceles and we have $AH=PH$. Analogously, we can prove that $AH=QH$. Therefore, $H$ is the circumcenter of the triangle $APQ$.

(Karl Czakler) ☐